A259236 Increasing sequence of numbers n such that the digits of n appear as a substring of the concatenation of the terms < n of the sequence, which is seeded with 1..9.

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 23, 34, 45, 56, 67, 78, 89, 91, 99, 122, 123, 199, 212, 221, 222, 223, 231, 232, 233, 234, 312, 319, 322, 323, 332, 333, 334, 343, 344, 345, 431, 433, 434, 443, 444, 445, 454, 455, 456, 543, 544, 545, 554, 555, 556, 565, 566, 567, 654, 655, 656, 665, 666, 667, 676, 677, 678, 765, 766, 767, 776, 777

Offset: 1

Author

Anthony Sand, Jun 22 2015

Keywords

• nonn
• base
• easy

Comments

In A048991, n is excluded if it appears in the concatenation of all earlier terms. This sequence applies the opposite criterion and excludes n if it does NOT appear in the concatenation of all earlier terms. For example, the sequences starts with 1, 2, 3, ..., therefore 12, 23 and 123 appear in the sequence, but 10, 11 and 14 do not.

Examples

			The digits of 12 appear earlier in the sequence (1, 2...), therefore 12 is included.
The digits of 11 do not appear earlier in the sequence, therefore 11 is excluded.
		

Links

• Anthony Sand, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A048991.

Programs

Maple

M:= 4: # to get all terms with <= M digits
with(StringTools):
S:= "123456789":
nS:= length(S):
Substrings:= {seq(seq(SubString(S,a..b),b=a+1..min(9,a+M-1)),a=1..8)}:
Cands:= map(parse,Substrings):
for n from 1 to 9 do A[n]:= n od:
for n from 10 while Cands <> {} do
  m:= min(Cands);
  A[n]:= m;
  S:= cat(S,convert(m,string));
  nm:= length(m);
  newSubstrings:= {seq(seq(SubString(S,a..b), b = a+nm-1..min(nS+nm, a+M-1)),a=1+nS-M .. nS)};
  Cands:= select(`>`,Cands union map(parse,newSubstrings), m);
  nS:= length(S);
od:
seq(A[i],i=1..n-1); # Robert Israel, Jun 22 2015

PARI

{ dmx=1000; d=vector(dmx); b=10; for(i=1,b-1,d[i]=i;print1(i,", ")); di=b-1; n=di; while(di
				

Formula

digits(n,i=1,j) = substring(sequence,i=1,j))